Answer
The area is bounded by
$$
\left( y=x^{3}(x^{2}-1)^{\frac{1}{3}}, \quad y =0, \quad x=1, \quad x=3 \right)
$$
is equal to
$$
\int_{0}^{1} x^{3}(x^{2}-1)^{\frac{1}{3}}dx=\frac{234}{7}\approx 33.43
$$
Work Step by Step
The area is bounded by
$$
\left( y=x^{3}(x^{2}-1)^{\frac{1}{3}}, \quad y =0, \quad x=1, \quad x=3 \right)
$$
is equal to
$$
\int_{0}^{1} x^{3}(x^{2}-1)^{\frac{1}{3}}dx
$$
First, we evaluate the indefinite integral as follows:
$$
\int x^{3}(x^{2}-1)^{\frac{1}{3}}dx
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=x^{2}, \quad\quad dv= x(x^{2}-1)^{\frac{1}{3}}dx } \\ {d u= 2xdx, \quad\quad v= \frac{3}{8}(x^{2}-1)^{\frac{4}{3}}dx }\end{array}\right] ,
$$
then
$$
\begin{aligned} \int x^{3}\left(x^{2}-1\right)^{1 / 3} d x&=\\
&=\frac{3 x^{2}}{8}\left(x^{2}-1\right)^{4 / 3}-\frac{3}{4} \int x\left(x^{2}-1\right)^{4 / 3} d x \\
&=\frac{3 x^{2}}{8}\left(x^{2}-1\right)^{4 / 3}-\frac{3}{4}\left[\frac{1}{2} \cdot \frac{3}{7}\left(x^{2}-1\right)^{7 / 3}\right] \\
&=\frac{3 x^{2}}{8}\left(x^{2}-1\right)^{4 / 3}-\frac{9}{56}\left(x^{2}-1\right)^{7 / 3}+C \\
\end{aligned}
$$
Now we will evaluate the definite integral as follows:
$$
\begin{aligned} A &=\int_{1}^{3} x^{3}(x^{2}-1)^{\frac{1}{3}}dx \\ &=\left.\left[\frac{3 x^{2}}{8}\left(x^{2}-1\right)^{4 / 3}-\frac{9}{56}\left(x^{2}-1\right)^{7 / 3}\right]\right|_{1} ^{3} \\ &=\frac{3}{8}(144)-\frac{9}{56}(128) \\ &=54-\frac{144}{7} \\
&=\frac{234}{7} \\
&\approx 33.43 \end{aligned}
$$
